Consider the curve represented parametrically by the equation `x = t^3-4t^2-3t` and `y = 2t^2 + 3t-5` where `t in R`.If H denotes the number of point on the curve where the tangent is horizontal and V the number of point where the tangent is vertical then

Options

A. H=2 and V=1
B. H=1 and V=2
C. H=2 and V=2
D. H=1 and V=1

Answer

To solve the problem, we need to find the number of points on the curve where the tangent is horizontal (H) and vertical (V). The curve is given parametrically by the equations:

\[ x = t^3 - 4t^2 - 3t \]
\[ y = 2t^2 + 3t - 5 \]

### Step 1: Find \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\)

First, we differentiate \(y\) and \(x\) with respect to \(t\):

1. **Differentiate \(y\)**:
\[ \frac{dy}{dt} = \frac{d}{dt}(2t^2 + 3t - 5) = 4t + 3 \]

2. **Differentiate \(x\)**:
\[ \frac{dx}{dt} = \frac{d}{dt}(t^3 - 4t^2 - 3t) = 3t^2 - 8t - 3 \]

### Step 2: Find the slope \(\frac{dy}{dx}\)

The slope of the tangent to the curve is given by:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{4t + 3}{3t^2 - 8t - 3} \]

### Step 3: Find points where the tangent is horizontal (H)

The tangent is horizontal when \(\frac{dy}{dx} = 0\). This occurs when the numerator is zero:
\[ 4t + 3 = 0 \]
Solving for \(t\):
\[ 4t = -3 \implies t = -\frac{3}{4} \]
Thus, there is **1 point** where the tangent is horizontal, so \(H = 1\).

### Step 4: Find points where the tangent is vertical (V)

The tangent is vertical when \(\frac{dy}{dx}\) is undefined, which occurs when the denominator is zero:
\[ 3t^2 - 8t - 3 = 0 \]
We will solve this quadratic equation using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \(a = 3\), \(b = -8\), and \(c = -3\):
\[ t = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 3 \cdot (-3)}}{2 \cdot 3} \]
Calculating the discriminant:
\[ (-8)^2 = 64 \quad \text{and} \quad 4 \cdot 3 \cdot (-3) = -36 \implies 64 + 36 = 100 \]
Now substituting back:
\[ t = \frac{8 \pm \sqrt{100}}{6} = \frac{8 \pm 10}{6} \]
Calculating the two possible values:
1. \(t = \frac{18}{6} = 3\)
2. \(t = \frac{-2}{6} = -\frac{1}{3}\)

Thus, there are **2 points** where the tangent is vertical, so \(V = 2\).

### Final Result

The values are:
- \(H = 1\)
- \(V = 2\)